Posted by Barry on October 18, 2002 at 22:22:53:
[t (radians) such that 0 < t < (1/2)(pi) ]
T(t) = (5 / x)[t + 2(cos t)]
dT/dt = (5 / x)[1 – 2(sin t)]
(5 / x)[1 – 2(sin t)] = 0
1 – 2(sin t) = 0
1 = 2(sin t)
sin t = 1/2
t = arcsin (1/2)
t = (1/6)(pi) radians or 30 degrees
Hence, (1/6)(pi) radians or 30 degrees.
Could someone explain in detail how the "derivative" is arrived at. Thank you.
I know the 1st derivative of (example) 3X^5 = 15x^4.